1. Thermodynamics

2. Energy: Ability to do work or produce heat. Work=force x distance force causes the object to move  Gravitational force causes the water to fall.  can generate electricity Energy kinetic potential  energy possessed by an object in virtue of its motion.  E kin =1/2 mv 2  E kin =3/2 RT Never confuse T and heat Heat is the energy transferred from one object to another in virtue of T-difference

3. Potential energy:  energy possessed by an object due to its presence in a force field i.e. under the effect of external force.  Object attracted/repelled by external force.  stored energy! E pot =mgh Attraction causes the ball to fall, h smaller, E pot smaller. Attraction causes the potential energy to decrease. Repulsion causes the potential energy to increase.

4. Law of conservation of Energy ( Axiome ): Energy can neither be created nor destroyed. Energy of universe is constant. Energy can be converted from one form to another. E kin ↔ E pot Heat ↔ Work Thermodynamics: the study of energy transformation from one form to another. First Law of TD.

5. System Part of universe under investigation. sys surroundings sys + surr = universe

7. State Function Change in state function depends only on initial and final state. Irbid Amman Sea level h=650 m h=900 m Irbid → Amman  h=h final -h initial  h=h amman -h irbid  h=900 m-650 m=250 m Initial state Final state Change doesn’t depend on path

8. Examples of state functions: –Temperature –Volume –Pressure –Altitude –Mass –Energy –Concentration

9. State function

10. (  U) x (  U) y

11. Schwartz theorem Equality of cross-derivatives

13. Show that p in the VdW equation is a state function!

14. Circular Rule

16. Internal Energy E Sum of E kin and E pot of all particles in the system. State function First Law of TD  E = Q + W The internal energy of a system can be changed 1.by gaining or losing heat, Q 2.Work, W, done on the system or by the System

17. Heat energy flow from one system to another in virtue of temperature difference. –Appears only during a change in state –Flows across the boundary from the a point of higher temperature to a point of lower temp –Not a state function –Path function

18. Work Any quantity of energy that “flows” across the boundary between the system and the surroundings as a result of force acting through a distance. –Appears only during a change in state –Completely convertible into lifting a weight in the surr –Not a state function –Path function

19.  E = Q + W electricity (work) can be completely converted into lifting of weight.

20. Q and w are path functions (Depend on path). full initial empty final Path 1 Path 2

21. sys surroundings Q heat transferred from surr to sys. Surr loses heat, loses E, E surr ↓ Sys gains heat, gains E, E sys  for Surr: Q < 0 (neg.),  E < 0 for Sys: Q > 0 (pos.),  E > 0 sys surroundings Q heat transferred from sys to surr. Sys loses heat, loses E, E sys ↓ Surr gains heat, gains E, E surr  for Sys: Q < 0 (neg.),  E < 0 for Surr: Q > 0 (pos.),  E > 0  E = Q + W Q sys > 0 : endothermic process Q sys < 0 : exothermic process  Q sys > 0  Q sys < 0 dE =  Q +  W

22. m1m1 sys surr m1m1 sys surr m2m2 m1m1 m2m2 sys E p =mgh h ↓, E p ↓ h of m 1 and m 2 ↓ E p of m 1 and m 2 ↓ E surr ↓ E sys  Work done by surroundings on system (E sys ) f > (E sys ) i  E sys > 0 w sys > 0

23. m1m1 sys surr m1m1 m2m2 sys E p =mgh h , E p  h of m 1  E p of m 1  E surr  E sys ↓ Work done by system on surroundings (E sys ) f < (E sys ) i  E sys < 0 w sys < 0 m1m1 surr sys

24. Ex. 6.1 A system undergoes an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ work is done on the system. Calculate the total change in the internal Energy of the system. Q sys > 0Q=+15.6 kJ w sys > 0w=+1.4 kJ  E sys = Q sys + w sys  E sys = (+15.6 kJ) + (+1.4 kJ) = +17 kJ

25. m1m1 sys surr hihi m1m1 sys hfhf initial final - m1m1 m2m2 surr sys

26. Ideal Gas One-stage isothermal Expansion T=constant

27. m1m1 sys surr hihi hfhf initial final - m1m1 m2m2 surr sys m1m1 m 2 -dm surr sys In each step: - dm is removed - Piston rises by dh - p sys = p opp - dw=-p opp dV - dw=-p sys dV

28. Reversible isothermal Expansion reversible: System in equilibrium with surroundings at each point of process Ideal gas

30. Reversible Expansion Ideal gas V f > V i w < 0 Reversible Compression Ideal gas V f < V i w > 0 Reversible work larger than irreversible work. But impractical. Requires infinity long time.

31. Ex. 6.2 Calculate the work associated with the expansion of a gas from 46 L to 64 L at constant external pressure of 15 atm. 46 L 64 L 15 atm o Expansion against the external pressure o External pressure opposes the expansion o p opp =15 atm = constant

32. Ex. 6.3 Given a balloon with a volume of 4.00x10 6 L. It was heated by 1.3x10 8 J until the volume became 4.5x10 6 L. Assuming the balloon is expanding against a constant external pressure of 1 atm, calculate the change in the internal energy of the gas confined by the balloon. 4.00x10 6 L 1 atm 4.50x10 6 L ViVi VfVf Q p opp

34. Heat Molar heat capacity Electrical work w=V.I.t → system gains this energy as heat q Kinetic energy of system rises → temperature rises Translational, rotational, vibrational energy increase

35. c p,m : molar heat capacity at constant pressure heat needed to raise the temperature of 1 mole of substance by 1ºC at constant pressure. c V,m : molar heat capacity at constant volume heat needed to raise the temperature of 1 mole of substance by 1ºC at constant volume.

36. c p : heat capacity at constant pressure heat needed to raise the temperature of n mole of substance by 1ºC at constant pressure. c V : heat capacity at constant volume heat needed to raise the temperature of n mole of substance by 1ºC at constant volume.

37. Extensive and intensive properties properties extensive intensive depends on amount of substance c p, V, m, n doesn’t depend on amount of substance c p,m, p, T Ratio of two extensive properties c p,m, Mwt,

39. Measure q V → change in internal energy of system  E determined

42. Enthalpy, H First law under the conditions -Reversible p opp =p sys -constant pressure p sys =const

51. Adiabatic compression: work done on system → internal energy increase → T↑ Adiabatic expansion: work done by system → internal energy decrease → T↓

52. reversible adiabatic work for ideal gas: much easier

53. Dependence of internal energy on volume changes For ideal gas: For VdW gas:

54. Two moles of a VdW gas were compressed at constant temperature of 25 o C from 10 L to 1 L. Calculate the change in the internal energy E.

55. Dependence of enthalpy on pressure changes

57. For ideal gas: For liquids and solids:

58. Relation between c p and c v

59. For ideal gas: For liquids and solids: because is rather small

62. Joule-Thompson Effect p 1 =constant p 2 =constant Insulated tube → q=0

63. isoenthalpic Joule-Thompson coefficient

64. For ideal gas: Pressure drop: p 1 > p 2 dp < 0 dT = 0 no change in Temp

65. For VdW gas:Can be shown at zero-pressure limit: J-T inversion temperature: T at which  J-T =0 Expansion at this Temp: no change in T

66. If a nitrogen gas cylinder is opened at room temp: Frost formation on valve If a hydrogen gas cylinder is opened at room temp: increase of valve temp eventual explosion

68. Liquefying Air

69. Chemical Energy CH 4(g) +2O 2(g) → CO 2(g) + 2H 2 O (g) C-HO=O C=O O-H Chemical reaction: o No change in the number/nature of atoms o Redistribution of Bonds (change in bonding) o Change in attraction & repulsion forces between the atoms o Change in the potential energy E p of molecules EpEp R P  Energy is conserved!  Energy difference released as heat.  Heat of reaction (Q v, Q p ).  Reaction exothermic.  H reaction = H f – H i = H P – H R < 0

70. N 2(g) + O 2(g) → 2NO (g) N≡NN≡NO=O N=O EpEp R P  Energy is conserved!  Energy difference obtained from surroundings as heat.  Heat of reaction (Q v, Q p =  E,  H).  Reaction endothermic.  H reaction = H f – H i = H P – H R > 0  E reaction = E f – E i = E P – E R > 0 o E p (R) > E p (P), reaction exothermic o E p (R) < E p (P), reaction endothermic

71. Thermochemical equation N 2(g) + O 2(g) → 2NO (g)  H o =+180.5 kJ CH 4(g) +2O 2(g) → CO 2(g) + 2H 2 O (g)  H o =-802.3 kJ moles 1 mole of gaseous methane (CH 4 ) reacts with two moles of gaseous molecular oxygen producing 1 mole of gaseous carbon dioxide, 2 moles of water vapor and 802.3 kJ of heat.  H o : Standard heat of reaction: Standard conditions: T=25 o C, p=1atm.

72. Calorimetry calorie cal measurement heat unit 1 cal = 4.185 J Calorimetry = heat measurement experiments 1 Cal =1000 cal Problem: - heat (Q) is a path function!!! - Q differs from one way of performing the experiment to another. - details of the experiment must be described!!!!

73. heat measurement experiments Heat measured at constant volume: o Equal to  E o Equal to a change in a state function!! o Details of the experiment no more important.

74. However, heat measurement experiments are usually performed at constant pressure Heat measured at constant pressure: o Equal to  H o Equal to a change in a state function!! o Details of the exp. no more important.

75. c sp : specific heat (specific heat capacity) heat needed to raise the temperature of 1 gram of substance by 1ºC. C : heat capacity heat needed to raise the temperature of substance (m gram) by 1ºC. T increase by 1ºC: 1 g c sp m gr. ? = C Q : heat heat needed to raise the temperature of substance (m gram) by a given temperature difference,  TºC. m gr. C T increase by 1ºC T increase by  TºC ? = Q

76. Bomb Calorimeter

77. 0.5269 g of octane (C 8 H 18 ) were placed in a bomb calorimeter with a heat capacity of 11.3 kJ/ºC. The octane sample was ignited in presence of excess oxygen. The temperature of the calorimeter was found to increase by 2.25ºC. Calculate  E of the combustion reaction of octane.  E defined for the reaction as written!!!!!!!!!!! C 8 H 18(g) +12.5O 2(g) → 8CO 2(g) + 9H 2 O (g)  E defined for the combustion of 1 mole octane (114.2 g)!! 0.5269 g QVQV 114.2 g ? =  E  E = - (114.2 g x 25.4 kJ )/0.5269 g =-5505 kJ

78. When 1.5 g of methane (CH 4 ) was ignited in a bomb calorimeter with 11.3 kJ/ºC heat capacity, the temperature rised by 7.3ºC. When 1.15 g hydrogen (H 2 ) was ignited in the same calorimeter, the temperature rised by 14.3ºC. Which one of the two substances has a higher specific heat of combustion (i.e. heat evolved upon the combustion of 1 g of substance)? 1.5 g Q V =83 kJ 1 g ? =55 kJ/g 1.15 g Q V =162 kJ 1 g ? =141 kJ/g

79. Coffee-Cup Calorimeter 50 mL of 1.0 M HCl at 25ºC were added to 50 mL of 1.0 M NaOH at 25ºC in a coffee-cup calorimeter. The tempe- rature was found to rise to 31.9ºC. Calculate the heat of the neutraliza- tion reaction! Was caused the temperature to increase? Exothermic Reaction HCl (aq) +NaOH (aq) → NaCl (aq) + H 2 O (l) H + (aq) +OH - (aq) → H 2 O (l)

80. heat evolved = heat gained + heat gained by reaction by solution by calorimeter Assumptions:  C cal =0 (very small mass)  Solution ≈ water  (c sp ) solution =(c sp ) water =4.18 Jg -1 ºC -1  (density) solution =(density) water =1 g/mL

81. HCl (aq) +NaOH (aq) → NaCl (aq) + H 2 O (l) n HCl =M HCl x V HCl = 1 mol/L x 0.050 L = 0.050 mol n NaOH =M NaOH x V NaOH = 1 mol/L x 0.050 L = 0.050 mol 0.050 mol 0.050 mol H 2 O2884.2 J mol 1 mol H 2 O ?Q p =57,684 J/mol H2O  H= - 57,684 J/mol H2O  H= - 57.7 kJ/mol H2O

82. Hess’s Law N 2(g) + 2O 2(g) initial 2NO 2(g) final 2NO (g) O 2(g) path 1 path 2

83. N 2(g) + O 2(g) → 2NO (g)  H 2a 2NO (g) + O 2(g) → 2NO 2(g)  H 2b N 2(g) + 2O 2(g) → 2NO 2(g)  H 1  H 1 =  H 2a +  H 2b The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps. If a chemical equation can be written as the sum of several other chemical equations (steps), the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations (steps). Hess’s Law:

84. Rules for manipulating thermochemical equations - If equation is multiplied by a factor, multiply  H by this factor. N 2(g) +3H 2(g) → 2NH 3(g)  H=-92 kJ 2 x (N 2(g) +3H 2(g) → 2NH 3(g)  H=-92 kJ) 2N 2(g) +6H 2(g) → 4NH 3(g)  H=-184 kJ 1/2 x (N 2(g) +3H 2(g) → 2NH 3(g)  H=-92 kJ) 1/2N 2(g) +3/2H 2(g) → NH 3(g)  H=-46 kJ - If equation is reversed, change the sign of  H 2NH 3(g) → N 2(g) + 3H 2(g)  H=+92 kJ

85. The enthalpy of combustion of graphite is -394 kJ/mol. The enthalpy of combustion of diamond is -396 kJ/mol. Calculate  H for the reaction: C graphite → C diamond Solving Strategy Write the given data in form of thermochemical equations: C G + O 2(g) → CO 2(g)  H=-394 kJ C D + O 2(g) → CO 2(g)  H=-396 kJ Construct the equation of interest from the given data:  1 mole c graphite is needed as reactant. Take the equation in the given data that contains c graphite. Check the number of moles and whether it is on the reactant side. Manipulate if necessary. C G + O 2(g) → CO 2(g)  H=-394 kJ  1 mole c diamond is needed as product. Take the equation in the given data that contains c diamond. Check the number of moles and whether it is on the product side. Manipulate if necessary. CO 2(g) → C D + O 2(g)  H=+396 kJ

86.  Sum the resulting equations and their  H values: C G + O 2(g) → CO 2(g)  H=-394 kJ CO 2(g) → C D + O 2(g)  H=+396 kJ C graphite → C diamond  H=+2 kJ

87. Given: 2 B (s) +3/2 O 2(g) → B 2 O 3(s)  H=-1273 kJ B 2 H 6(g) +3 O 2(g) → B 2 O 3(s) + 3 H 2 O (g)  H=-2035 kJ H 2(g) +1/2 O 2(g) → H 2 O (l)  H=-286 kJ H 2 O (l) → H 2 O (g)  H=+44 kJ Calculate  H for 2 B (s) + 3 H 2(g) → B 2 H 6(g) 2 B (s) +3/2 O 2(g) → B 2 O 3(s)  H=-1273 kJ B 2 O 3(s) + 3 H 2 O (g) → B 2 H 6(g) +3 O 2(g)  H=+2035 kJ 3H 2(g) +3/2 O 2(g) → 3 H 2 O (l)  H=3 x (- 286) kJ 2 B (s) + 3 H 2 O (g) + 3 H 2(g) → B 2 H 6(g) +3 H 2 O (l)  H=-96 kJ 3 H 2 O (l) → 3 H 2 O (g)  H=3 x (+44) kJ 2 B (s) + 3 H 2(g) → B 2 H 6(g  H=+36 kJ

88. Heat of Formation Formation reaction: reaction of forming 1 mole of product from the elements in their stable form at 25ºC and 1 atm. Heat of formation =  H of formation reaction =  F H Standard heat of formation =  Hº of formation reaction =  F Hº  F Hº(NO (g) ): ½ N 2(g) +½ O 2(g) → NO (g)  Hº  F Hº(CO (g) ): C graphite(s) +½ O 2(g) → CO (g)  Hº  F Hº(O (g) ): ½ O 2(g) → O (g)  Hº  F Hº(C diamond(s) ): C graphite(s) → C diamond(s)  Hº  F Hº(O 2(g) ): O 2(g) → O 2(g)  Hº=0  F Hº(C graphite(s) ): C graphite(s) → C graphite(s)  Hº=0

90. CH 4(g) +2O 2(g) → CO 2(g) + 2H 2 O (g)  H=? C G(s) + O 2(g) → CO 2(g)  F H(CO 2 ) 2 x ( H 2(g) +1/2 O 2(g) → H 2 O (g) ) 2 x  F H(H 2 O) CH 4(g) → C G(s) + 2 H 2(g) -  F H(CH 4 ) O 2(g) → O 2(g) -  F H(O 2 )=0 CH 4(g) +2O 2(g) → CO 2(g) + 2H 2 O (g)  H=  F H(CO 2 )+ 2 x  F H(H 2 O) -  F H(CH 4 ) -  F H(O 2 )  H=  F H(CO 2 )+ 2 x  F H(H 2 O) – [  F H(CH 4 ) +  F H(O 2 )]

91. 4 NH 3(g) +7 O 2(g) → 4 NO 2(g) + 6 H 2 O (l)  H=?  H= 4 x  F H(NO 2 )+ 6 x  F H(H 2 O) – 4 x  F H(NH 3 ) 2 Al (s) +Fe 2 O 3(s) → Al 2 O 3(s) + 2 Fe (s)  H=?  H=  F H(Al 2 O 3 )+ 2 x  F H(Fe) – [  F H(Fe 2 O 3 )+ 2 x  F H(Al)]  H=  F H(Al 2 O 3 ) –  F H(Fe 2 O 3 )

92. 2 CH 3 OH (l) +3 O 2(g) → 2 CO 2(g) + 4 H 2 O (l)  H=?  H= 2 x  F H(CO 2 )+ 4 x  F H(H 2 O) – 2 x  F H(CH 3 OH)  H= 2 x (-394 kJ)+ 4 x (-286 kJ) – 2 x (-239 kJ)=-1454 kJ 2 mol CH 3 OH-1454 kJ 2x32 g ? -1454 kJ 1 g = -22.7 kJ/g Calculate the heat of combustion of methanol (CH 3 OH (l) ) in kJ/g and compare its value with that of octane (C 8 H 18(l) ).

93. C 8 H 18(l) +12.5 O 2(g) → 8 CO 2(g) + 9 H 2 O (l)  H= 8 x  F H(CO 2 )+ 9 x  F H(H 2 O) –  F H(C 8 H 18 )  H= 8 x (-394 kJ)+ 9 x (-286 kJ) – (-276 kJ)=-5450 kJ 1 mol C 8 H 18 -5450 kJ 114 g ? -5450 kJ 1 g = -47.8 kJ/g